Chapter 14, Problem 14.25P

Interpretation Introduction

Interpretation:

Cell voltage when value of $\left[{\text{Fe}}^{\text{2+}}\right]\text{/}\left[{\text{Fe}}^{\text{3+}}\right]$ is $2.5\times {10}^{-3}$ has to be calculated.

Concept Introduction:

Cell that is more negatively charged has better oxidation tendency and thus will readily lose electron. Electrons released from this electrode will reach positively charged reduction potential electrode.

For a general representation, the oxidation half cell is drawn in left separated by double vertical parallel lines while reduction half cell comes at right.

Expression to compute $E_{\text{cell}}$ as per the Nernst equation is written as follows:

  $E_{\text{cell}}=E_{\text{cell}}^{°}-\frac{0.0592}{n}\log \frac{\left[P\right]}{\left[R\right]}$

Here,

• $E_{\text{cell}}$ denotes overall cell potential.
• $E_{\text{cell}}^{°}$ denotes standard cell potential.
• $n$ is moles of electrons transferred in each reaction.
• $\left[P\right]$ denotes concentration of products.
• $\left[R\right]$ denotes concentration of reactant.